algebraic manipulation – Rearranging inequality

I’m trying to rearrange inequalities by collecting the together terms that involve the same powers of a variable of my choice. Here is an example:

$$a + b a – 1 > a^2 – c a – b c$$

I’d like to specify $a$ as the variable and rearrange it as

$$-a^2 + (b+c+1)a>1-bc quad text{or} quad -a^2 + (b+c+1)a-1+bc>0$$

I’ve tried to achieve that with Collect, but couldn’t make it work on both sides of the inequality. As noted by a commenter Simplify may help in this particular example. However, in general, it won’t pick the common terms for my variable of interest.

Any ideas on how to achieve that?

algebraic topology – Identification of product of two spheres from wedge of two spheres

$mathbf {The Problem is}:$ Show that $S^2×S^2$ can be obtained by attaching a $4-$cell to $W=S^2vee S^2.$

$mathbf {My approach}:$ Actually, I am a beginner in learning CW complexes .
I was thinking to draw a pushout such that $S^3$ is attached to $W$ at the wedge point $*.$

But, I can’t imagine properly , somehow; I am trying to glue $S^3$ to previous skeleton .

A small hint with a motivation in solving these type of problems is warmly appreciated, thanks in advance .

ag.algebraic geometry – Base change to algebraic closure commutes with quotient of polynomial ring by maximal ideal

Let $k$ be a field, $R:=k(x_1, cdots , x_n)$ and $mathfrak m$ be a maximal ideal such that $R/mathfrak m$ is a finite separable field extension of $k$. Consider the algebraic closure $overline k$ of $k$ and let $mathfrak m_1, dots , mathfrak m_r$ be the maximal ideals of the ring $$overline R :=overline k otimes_k R = overline k(x_1, cdots , x_n)$$ lying over $mathfrak m$.

In an article I have been reading, the following isomorphism is used:
$$overline k otimes_k R/mathfrak m cong overline R big/ bigcap_{1 leq j leq r} mathfrak m_j$$

Since I have not been able to find any reference for this, I have trying to justify the same myself:

In the forward direction, I could make the usual map work, namely, the one sending each element $$(alpha, f + mathfrak m) in overline k times R/mathfrak m text{ (for }f in Rtext{) }hspace{2mm}text{ to }hspace{2mm} alpha f+ bigcap_{i=1}^r mathfrak m_i in R big/ bigcap_{1 leq j leq r} mathfrak m_j.$$
Indeed it turns out to be well-defined since each $mathfrak m_i$ lies over $mathfrak m$ and $k$-bilinearity is checked easily, whereupon the universal property of the tensor product applies.

However it is while showing the inverse isomorphism that I have gotten stuck. My idea was to try and get a map
$$phi: overline R longrightarrow overline k otimes_k R/mathfrak m$$
whose kernel contains the intersection $bigcap_{j=1}^r mathfrak m_j$, and to do the same I tried sending each $alpha otimes f in overline k otimes_k R = overline R$ to $alpha otimes overline f = alpha otimes (f + mathfrak m) in overline k otimes_k R/mathfrak m$. However, in order to be able to use the Mapping Property and get my desired map $phi$, I seem to need to have $bigcap_{i=1}^r mathfrak m_i subset mathfrak m$, which in this case becomes equivalent to $bigcap_{i=1}^r mathfrak m_i = mathfrak m$).

And it is not clear to me why this last equality should hold; perhaps I am missing something obvious or some property of the ring extension $R subset overline R$ is at work here. I tried to use the weak Nullstellensatz to note that each $mathfrak m_i$ must be of the form $(x_1-a_{i1}, cdots , x_n-a_{in})$ for some $(a_{i1}, cdots , a_{in}) in overline k^n$, but this hasn’t proven to be useful yet. I am not sure whether the condition of $R/mathfrak m$ being separable over $k$ will be helpful here.

I would really appreciate a proof (perhaps one using the above ideas?) or reference. Thank you.

algebra precalculus – Algebraic Manipulation Question When Manipulating Minimum in Epsilon Delta Proof

I am currently working my way through Spivak’s Calculus, and I cannot figure out the last basic algebra step to reduce the problem so that it reads as presented in the book’s solutions.

Specifically, I am asked to find a $delta$ such that $lvert{x^4 – a^4}rvert < epsilon ; forall x$ satisfying $lvert{x-a}rvert < delta$.

The way to tackle the problem is very straightforward. Notice that $lvert{x^2 – a^2}rvert < min{big(frac{epsilon}{2(lvert{a^2}rvert + 1)}, 1big)} implies lvert{x^4 – a^4}rvert < epsilon$. Furthermore $lvert{x – a}rvert < min{big(frac{epsilon’}{2(lvert{a}rvert + 1)}, 1big)} implies lvert{x^2 – a^2}rvert < epsilon’$. Substituting $epsilon = epsilon’$, we have:

$$
min{bigg(frac{min{big(frac{epsilon}{2(lvert{a^2}rvert + 1)}, 1big)}}{2(lvert{a}rvert + 1)}, 1bigg)}
$$
.

After I simplify, I choose:

$$
delta = min{bigg(min{big(frac{epsilon}{4(lvert{a^2}rvert + 1)(lvert{a}rvert + 1)}, frac{1}{2(lvert{a}rvert + 1)}big)}, 1bigg)}
$$

My Question:

Spivak’s solutions show that this may be reduced to

$$
delta = min{big(frac{epsilon}{4(lvert{a^2}rvert + 1)(lvert{a}rvert + 1)}, 1big)}
$$

and I cannot figure out why this is true. Why can we guarantee that

$$

begin{align}
frac{epsilon}{4(lvert{a^2}rvert + 1)(lvert{a}rvert + 1)} &< frac{1}{2(lvert{a}rvert + 1)} \
frac{epsilon}{2(lvert{a^2}rvert + 1)} &< 1
end{align}

$$

for any such $epsilon$?

algebraic geometry – Constant sheaf with value the empty set

Let X be a topological space. Is there such a thing as the constant sheaf on X with value the empty set? That is, a sheaf that assigns to each open set U the set of maps from U to the empty set and to each inclusion the empty map?

I ask because I am trying to show that the projection map from the etale space of a locally constant sheaf on X to X, defined to send a germ at a point x to the point x, is a covering map. It’s not clear to me that this map is surjective. That is, why can’t there be an x in X such that there are no sections over any neighborhood of x?

ag.algebraic geometry – Example of an Algebraic Space (“false” affine line with different tangents at origin)

I have a question about following example from the Intro
on algebraic spaces
by Roy Mikael Skjelnes (page 12)
of a presheaf quotient, which
has associated sheaf which can not be a scheme.
It’s originally an example from Knutson’s book on Algebraic spaces, but there my concern also isn’t
answered. Here the example:

enter image description here

By consruction of the quotient of $R rightrightarrows X$
the resulting presheaf quotient has, over the origin, two
different tangent directions sticking out.

Then it is clamed that since
presheaf quotient is separated (Proposition
(1.13)), these tangent directions will also appear
in the sheaf quotient. I not understand. The Proposition
1.13 states:

Proposition 1.13. Let $Cov$ be a pretopology, and let $R$ and
$X$ be two sheaves. Assume that we have an equivalence relation
$R rightrightarrows X$, and
let $X_R$ denote its presheaf quotient. Then the presheaf $X_R$
is separated.
In particular we have that $LX_R$ is a sheaf, and that the equivalence
relation is effective, i.e. $R = X times{LX_R} X$.
(here: $LF$ is def by
$LF(S):= lim_{T to S in Cov(S)} F(T)$ see also page 5)

Question: Proposition 1.13 states that only that the presheaf
(of the quotient) is separated, there is no statement about the separateness of the sheaf itself. So I not see why all tangent directions, which apear in the presheaf, must also apear without any loss in hypothetically existing associated sheaf?

Secondly: I know that separatedness means that it ‘separates’ points
(= $Spec(k) to X$-morphisms)
in algebro-geometric setting. Does it it here also ‘separates’
tangent directions?

ag.algebraic geometry – Generalization of a standard algebraic group theory result for a tensor problem

Let $x, y, z, a, b, c$ be positive integers with $x geq a$ and $y geq b$ and $z geq c$. Let $V=mathbb{C}^x otimes mathbb{C}^y otimes mathbb{C}^z$ be a tensor product space acted on algebraically by $G:=GL(mathbb{C}^x)times GL(mathbb{C}^y)times GL(mathbb{C}^z)$ in the standard way. Let $W subseteq V$ be the linear subspace defined by
begin{align}
W=text{span}{e_i otimes e_j otimes e_k : iin{1,dots, a}, jin{1,dots,b} kin {1,dots,c}},
end{align}

where $e_i, e_j, e_k$ are the standard basis vectors. For each $l in mathbb{N}$, define

begin{align}tag{1}
S_l={v in V : dim(overline{G cdot v} cap
W)leq l}.
end{align}

Is it true that $S_l subseteq V$ is Zariski closed for all $l in mathbb{N}$? If this statement is true, I would prefer a proof that uses classical algebraic group theory/algebraic geometry, as I am not very familiar with sheaves and schemes.

Note that if we remove $W$ from (1), then this becomes a very standard result in algebraic group theory (see e.g. Lemma 1.4 here).

algebraic manipulation – Why isn’t SameQ working as expected?

First time asking a question on here so forgive my format. Basically, I’m fairly new to Mathematica and I don’t understand why SameQ isn’t working like it should. If I try

Sqrt[1/x] === 1/Sqrt[x]

and evaluate the cell, it returns False. What exactly is going on here? I tried using FullSimply, Simplify, and EqualTo in various combinations but nothing seems to work.

number theory – An Algebraic Approach to the Goldbach Conjecture?

I have been thinking about the Goldbach Conjecture, (G.C.), off and on for the last few years and decided to explore using an algebraic method of proof. This has the benefits, as you will see, that very little needs to be known about the distribution of the prime numbers in order to prove the conjecture is true. I will put all lemmas, corollaries, and theorems in bold for easy reading.

First, it is important to state that the Goldbach Conjecture says that every even number greater than 2 can be written as the sum of two prime numbers. The goal is to determine algebraically what conditions are needed for the Goldbach Conjecture to fail and show those conditions cannot be met. In order to proceed it is important to note that throughout this proof the variable, $a$, will always be assumed to be a natural number greater than 3 and non-prime. Also, the set of natural numbers, $mathbb{N}$, will include zero. $mathbb{P}$ will represent the set of prime numbers. With this in place everything is needed to begin.

If the Goldbach Conjecture were false for some non-prime $a in mathbb{N}$ with $a > 3$, then the closure property of the integers ensures for any prime $p_i < a$ there exists some $q_i in mathbb{N}$ where
$$2a = q_i + p_i tag{Equation 1}$$
with no $q_i$ being prime. With $a < forall q_i < 2a$, there must exist $alpha_1, alpha_2, dots, alpha_{pi(a)} in mathbb{N}$ allowing for
$$prod_{i = 1}^{pi(a)}q_i = prod_{i = 1}^{pi(a)}p_i^{alpha_i}. tag{Equation 2}$$ Conversely, if for each prime $p_i < a$ there is some $q_i in mathbb{N}$ where $2a = q_i + p_i$ and there exists $alpha_1, alpha_2, dots, alpha_{pi(a)} in mathbb{N}$ allowing for $prod_{i = 1}^{pi(a)}q_i = prod_{i = 1}^{pi(a)}p_i^{alpha_i}$, then no $q_i$ can be prime. This would provide a counter-example to the Goldbach Conjecture.

Therefore, following lemma is true.

$$textbf{The G.C. is false only if there exists some $a in mathbb{N}$ where $a > 3$ and Equations 1, 2 are satisfied.} tag{Lemma 1}$$

The next step is to try and find solutions to these sets of equations. Notice from Equation 1 that it follows for any primes $p_i, p_j < a$ there must exist non-prime $q_i, q_j$ where
$$q_j + p_j = 2a = q_i + p_i. tag{Equation 3}$$

If some $p_i$ existed where $p_i | q_i$ and $p_i | q_j$ in Equation 3, then $p_i | p_j$. Since both $p_i, p_j in mathbb{P}$, it is impossible for $p_i | p_j$ when $j neq i$. Therefore, an important corollary can be produced
$$textbf{if any $p_i | q_i$, then $p_i nmid q_j$ when $j neq i.$} tag{Corollary 1}$$
The crucial final step is to show that it is impossible for any $p_i|q_j$ when $j neq i.$ This will be done by asking about the truth value of the inverse of Corollary 1. This is given by the statement if $p_i nmid q_i$, then $p_i|q_j$ for some $jneq i$.

To begin, notice that under Equation 2, if any $p_i | q_j$ for some $j neq i$, then $p_i|p_i^{alpha_i}$. This allows for another corollary given below.

$$textbf{If any $p_i|q_j$ for some $j neq i$, then $alpha_i > 0.$} tag{Corollay 2}$$

It will now be shown that the inverse of Corollary 1 is never true when its antecedent is true. To begin, take the conjunction of the inverse of Corollary 1 and Corollary 2 below.

$$text{(if $p_i nmid q_i$, then $p_i|q_j$ for some $jneq i$)$; land ;$ (if any $p_i|q_j$ for some $j neq i$, then $alpha_i > 0)$}$$
The conjunction above allows for the following conclusion via transitivity.
$$text{if $p_inmid q_i$, then $alpha_i > 0.$}$$
The law of contraposition states that the truth value of a conditional shares the same truth value of its contrapositive, therefore, it follows below.
$$text{(if $p_inmid q_i$, then $alpha_i > 0.)$ $; equiv ;$ (if $alpha_i = 0$, then $p_i|q_i$}.)$$
However, it is impossible for $alpha_i = 0$ and $p_i|q_i$ since this would require the two equal products in Equation 2 to have differing prime factors. This leads to a contradiction, as it violates the Fundamental Theorem of Arithmetic. This means that if $p_inmid q_i$, then $alpha_i > 0$ is false whenever $p_inmid q_i$ which demonstrates the inverse of Corollary 1 must also be false whenever $p_i nmid q_i$. Since either $p_i$ is a divisor of $q_i$ or not, along with the truth of Corollary 1, there exists no scenario where $p_i | q_j$ when $j neq i.$

Since it is impossible for any $p_i | q_j$ when $j neq i$ solutions to Lemma 1 must be of the form $q_i = p_i^{alpha_i}$. A substitution into Equation 1 shows
$$text{$2a = p_i^{alpha_i} + p_i$ for all prime $p_i < a$}tag{Equation 4}$$
forcing all $p_i > a$ to be divisors of $2a$. Bertrand’s Postulate ensures $2a < a#$ for $a > 4$, showing solutions cannot exist when $a > 3$. Under Lemma 1 the Goldbach Conjecture is true.

A similar procedure can be used to show that every even number is the difference of two prime numbers. Also, using the weak version of the Goldbach Conjecture along with everything proven thus far, it is possible to prove the Polignac Conjecture. I do not wish to go over all of those steps here, so a link to my paper is given by https://osf.io/terdc/download​. I have not been able to put this on the ARXIV since I have no university affiliation, and am honestly worrying about someone stealing the idea. At this point I have no other way to get this out to see if it has any merit and I have been sitting on it for years now hoping there would be another way. The journals I have submitted to have rejected it without any cause about 20 minutes after submitting it. If there is a mistake then I will accept that, but I cannot seem to find any. Thank you for any time that can be given. My email is in the paper.

algebraic number theory – Float point arithmetic

I got a simple problem for university where i got kind of stuck.
The question is following. Let $M>0$ be a signicand, B a unknown base and E an exponant so that $M*B^E$ is a valid number with t positions. We call it $M*B^Ein mathbb{F}_{B,t}$.
Furthermore we define $mathbb{F}_{B,t}={M*B^E: M=0 lor B^{t-1}leq |M|leq B^t}$. I now want to show, that if those assumptions are true, it follows, that $(M+1)*B^Ein mathbb{F}_{B,t}$.

Since $M*B^Ein mathbb{F}_{B,t}$ and $M>0$ it follows, that $B^{t-1}leq |M|< B^t$

My thought was that we make two cases:

First Case $Mneq B^{t}-1$: In this Case we can easily say, that it follows $B^{t-1}leq |M|leq |(M+1)|<B^t$ since we know that $B^{t-1}leq |M|< B^t-1$

Second Case: $M=B^{t}-1$: This is the point i am struggling with

My Problem is, that i do not find a way where i can keep the same exponent without changing the amount of positions in my significand. I either need to change t or E in my opinion, which makes me confused since i am not allowed to do either of those moves.